if ` 2A +B=[{:(5,-1),(3,2):}]and A-2B =[{:(1,-4),(0,5):}]` then find the matrices A and B .

Explanation

To solve the problem, we need to find the matrices \( A \) and \( B \) given the equations:

1. \( 2A + B = \begin{pmatrix} 5 & -1 \\ 3 & 2 \end{pmatrix} \) (Equation 1)
2. \( A - 2B = \begin{pmatrix} 1 & -4 \\ 0 & 5 \end{pmatrix} \) (Equation 2)

### Step 1: Rewrite the equations
Let's denote the matrices:
- \( A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \)
- \( B = \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} \)

From Equation 1:
\[ 2A + B = \begin{pmatrix} 5 & -1 \\ 3 & 2 \end{pmatrix} \]
This gives us:
\[ B = \begin{pmatrix} 5 & -1 \\ 3 & 2 \end{pmatrix} - 2A \]

From Equation 2:
\[ A - 2B = \begin{pmatrix} 1 & -4 \\ 0 & 5 \end{pmatrix} \]
This gives us:
\[ A = \begin{pmatrix} 1 & -4 \\ 0 & 5 \end{pmatrix} + 2B \]

### Step 2: Substitute for B
Substituting the expression for \( B \) from Equation 1 into the expression for \( A \):
\[ A = \begin{pmatrix} 1 & -4 \\ 0 & 5 \end{pmatrix} + 2\left(\begin{pmatrix} 5 & -1 \\ 3 & 2 \end{pmatrix} - 2A\right) \]
Expanding this, we have:
\[ A = \begin{pmatrix} 1 & -4 \\ 0 & 5 \end{pmatrix} + \begin{pmatrix} 10 & -2 \\ 6 & 4 \end{pmatrix} - 4A \]
Combining the terms gives:
\[ A + 4A = \begin{pmatrix} 1 + 10 & -4 - 2 \\ 0 + 6 & 5 + 4 \end{pmatrix} \]
\[ 5A = \begin{pmatrix} 11 & -6 \\ 6 & 9 \end{pmatrix} \]
Now, divide both sides by 5:
\[ A = \begin{pmatrix} \frac{11}{5} & -\frac{6}{5} \\ \frac{6}{5} & \frac{9}{5} \end{pmatrix} \]

### Step 3: Substitute A back to find B
Now substituting \( A \) back into the expression for \( B \):
\[ B = \begin{pmatrix} 5 & -1 \\ 3 & 2 \end{pmatrix} - 2\begin{pmatrix} \frac{11}{5} & -\frac{6}{5} \\ \frac{6}{5} & \frac{9}{5} \end{pmatrix} \]
Calculating \( 2A \):
\[ 2A = \begin{pmatrix} \frac{22}{5} & -\frac{12}{5} \\ \frac{12}{5} & \frac{18}{5} \end{pmatrix} \]
Now substituting this into the expression for \( B \):
\[ B = \begin{pmatrix} 5 & -1 \\ 3 & 2 \end{pmatrix} - \begin{pmatrix} \frac{22}{5} & -\frac{12}{5} \\ \frac{12}{5} & \frac{18}{5} \end{pmatrix} \]
Calculating the elements:
\[ B = \begin{pmatrix} 5 - \frac{22}{5} & -1 + \frac{12}{5} \\ 3 - \frac{12}{5} & 2 - \frac{18}{5} \end{pmatrix} \]
This simplifies to:
\[ B = \begin{pmatrix} \frac{25}{5} - \frac{22}{5} & -\frac{5}{5} + \frac{12}{5} \\ \frac{15}{5} - \frac{12}{5} & \frac{10}{5} - \frac{18}{5} \end{pmatrix} \]
\[ B = \begin{pmatrix} \frac{3}{5} & \frac{7}{5} \\ \frac{3}{5} & -\frac{8}{5} \end{pmatrix} \]

### Final Result
Thus, we have:
\[ A = \begin{pmatrix} \frac{11}{5} & -\frac{6}{5} \\ \frac{6}{5} & \frac{9}{5} \end{pmatrix}, \quad B = \begin{pmatrix} \frac{3}{5} & \frac{7}{5} \\ \frac{3}{5} & -\frac{8}{5} \end{pmatrix} \]